Evaluate $\int_{0}^{2} e^{x} dx$ as the limit of a sum.

(A) By definition,the definite integral as the limit of a sum is given by $\int_{a}^{b} f(x) dx = (b-a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + r\frac{b-a}{n})$.
Here,$a=0, b=2, f(x)=e^x$,so $\int_{0}^{2} e^x dx = (2-0) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} e^{0 + r\frac{2}{n}} = 2 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} (e^{2/n})^r$.
This is a geometric progression with $n$ terms,first term $1$,and common ratio $e^{2/n}$.
The sum is $\frac{1( (e^{2/n})^n - 1 )}{e^{2/n} - 1} = \frac{e^2 - 1}{e^{2/n} - 1}$.
Thus,$\int_{0}^{2} e^x dx = 2 \lim_{n \to \infty} \frac{1}{n} \left( \frac{e^2 - 1}{e^{2/n} - 1} \right) = 2(e^2 - 1) \lim_{n \to \infty} \frac{1/n}{e^{2/n} - 1}$.
Multiplying numerator and denominator by $2$,we get $2(e^2 - 1) \lim_{n \to \infty} \frac{2/n}{2(e^{2/n} - 1)} = (e^2 - 1) \lim_{h \to 0} \frac{h}{e^h - 1}$ where $h = 2/n$.
Since $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$,the result is $e^2 - 1$.